Question

Identify the reaction that cannot proceed through an inner-sphere electron transfer mechanism. For the other three, predict all possible six-coordinate products for inner-sphere electron transfer. Assume that each reaction occurs in water and that the second reagent is the oxidant in each case:

A. [Fe(H₂O)6]²⁺ + [Ru(NH₃)₅Cl]²⁺
B. [IrCl₄] + [Co(H₂O)₆]³⁺
C. [Co(H₂O)₆]²⁺ + [Cr(H₂O)₅Cl]²⁺
D. [Fe(H₂O)₆]²⁺ + [Fe(H₂O)₅Cl]²⁺

Answer 1

Option B, [IrCl₄] minus + [Co(H₂O)₆]³⁺, cannot proceed via inner-sphere electron transfer because it lacks an appropriate bridging ligand. The other reactions, A, C, and D, can form bridged bimetallic complexes with chlorine ligands displaced in the process.

Step-by-step explanation:

The reaction that cannot proceed through an inner-sphere electron transfer mechanism is option B, [IrCl₄] minus + [Co(H₂O)₆]³⁺, because it involves a tetrachloroiridate(II) complex which does not have an aquo or ammine ligand in its coordination sphere available to form a bridging ligand with the cobalt complex.

For the other reactions, the inner-sphere electron transfer can proceed as follows:

• A. [Fe(H₂O)₆]²⁺ + [Ru(NH₃)₅Cl]²⁺ → [Fe(H₂O)₅Ru(NH₃)₅]²⁸ + 2Cl⁺
• C. [Co(H₂O)₆]²⁺ + [Cr(H₂O)₅Cl]²⁺ → [Co(H₂O)₅Cr(H₂O)₅]´⁺ + 2Cl⁺
• D. [Fe(H₂O)₆]²⁺ + [Fe(H₂O)₅Cl]²⁺ → [Fe(H₂O)₅Fe(H₂O)₅]²⁴ + 2Cl⁺

The possible six-coordinate products involve the displacement of one water ligand for the formation of a bridging ligand between the two metal centers.